Question : Two circles having radii of $r$ units intersect each other in such a way that each of them passes through the centre of the other. Then the length of their common chord is:
Option 1: $\sqrt{2}r\ \text{units}$
Option 2: $\sqrt{3}r\ \text{units}$
Option 3: $\sqrt{5}r\ \text{units}$
Option 4: $r\ \text{units}$
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Correct Answer: $\sqrt{3}r\ \text{units}$
Solution :
Let $r$ be the radius of each circle.
AB is the common chord.
So, OA = OC = $r$
OE = $\frac{1}{2}×$OC = $\frac{1}{2}×r$
Using Pythagoras theorem on $\triangle$AOE, we get,
OA
2
= OE
2
+AE
2
⇒ AE = $\sqrt{r^2-(\frac{r}{2})^2}=\frac{\sqrt{3}r}{2}$
AB = 2 × AE = $2 × \frac{\sqrt{3}r}{2} = \sqrt{3}r\ \text{units}$
Hence, the correct answer is $\sqrt{3}r\ \text{units}$.
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